$(a)\;2.4\;mA \\(b)\;0.24\;mA \\ (c)\;24\;mA \\ (d)\;120\;mA $

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Final position after the coil has turned $180^{\circ}$ is

Initially $\theta=0$ and final $\theta= 180^{\circ}$

$\phi_1=NBA$

$\phi_2=-NBA$

$e_{av}=\large\frac{2NBA}{\Delta t}=\frac{2 \times 1200 \times 4 \times 10^{-4} \times 500 \times 10^{-4}}{0.1}$

$\qquad=0.48\;V$

$I =\large\frac{E_{aV}}{R} =\frac{0.48}{20}$

$\qquad= 24 \times 10^{-3}A$

Hence c is the correct answer.

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